Упр.20.1 ГДЗ Мордкович 10-11 класс (Алгебра)

1. Вычислите: а) $$\operatorname{tg}\frac{\pi}{12}$$; б) $$\operatorname{tg}105$$; в) $$\operatorname{tg}\frac{5\pi}{12}$$; г) $$\operatorname{tg}165$$.

Воспользуемся формулами тангенса суммы и разности:

$$\tg(\alpha \pm \beta)=\frac{\tg \alpha \pm \tg \beta}{1 \mp \tg \alpha \tg \beta}$$

1. $$\tg \frac{\pi}{12}=\tg\left(\frac{\pi}{4}-\frac{\pi}{6}\right)=\frac{\tg \frac{\pi}{4}-\tg \frac{\pi}{6}}{1+\tg \frac{\pi}{4}\tg \frac{\pi}{6}}=\frac{1-\frac{\sqrt{3}}{3}}{1+\frac{\sqrt{3}}{3}}=\frac{3-\sqrt{3}}{3+\sqrt{3}}$$

   $$\frac{3-\sqrt{3}}{3+\sqrt{3}}=\frac{(3-\sqrt{3})^2}{(3+\sqrt{3})(3-\sqrt{3})}=\frac{9-6\sqrt{3}+3}{9-3}=\frac{12-6\sqrt{3}}{6}=2-\sqrt{3}$$

2. $$\tg 105^\circ=\tg(45^\circ+60^\circ)=\frac{\tg 45^\circ+\tg 60^\circ}{1-\tg 45^\circ \tg 60^\circ}=\frac{1+\sqrt{3}}{1-\sqrt{3}}$$

   $$\frac{1+\sqrt{3}}{1-\sqrt{3}}=\frac{(1+\sqrt{3})^2}{(1-\sqrt{3})(1+\sqrt{3})}=\frac{1+2\sqrt{3}+3}{1-3}=\frac{4+2\sqrt{3}}{-2}=-2-\sqrt{3}$$

3. $$\tg \frac{5\pi}{12}=\tg\left(\frac{\pi}{4}+\frac{\pi}{6}\right)=\frac{\tg \frac{\pi}{4}+\tg \frac{\pi}{6}}{1-\tg \frac{\pi}{4}\tg \frac{\pi}{6}}=\frac{1+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}}=\frac{3+\sqrt{3}}{3-\sqrt{3}}$$

   $$\frac{3+\sqrt{3}}{3-\sqrt{3}}=\frac{(3+\sqrt{3})^2}{(3-\sqrt{3})(3+\sqrt{3})}=\frac{9+6\sqrt{3}+3}{9-3}=\frac{12+6\sqrt{3}}{6}=2+\sqrt{3}$$

4. $$\tg 165^\circ=\tg(45^\circ+120^\circ)=\frac{\tg 45^\circ+\tg 120^\circ}{1-\tg 45^\circ \tg 120^\circ}=\frac{1-\sqrt{3}}{1+\sqrt{3}}$$

   $$\frac{1-\sqrt{3}}{1+\sqrt{3}}=\frac{(1-\sqrt{3})^2}{(1+\sqrt{3})(1-\sqrt{3})}=\frac{1-2\sqrt{3}+3}{1-3}=\frac{4-2\sqrt{3}}{-2}=\sqrt{3}-2$$

Ответ: а) $$2-\sqrt{3}$$; б) $$-2-\sqrt{3}$$; в) $$2+\sqrt{3}$$; г) $$\sqrt{3}-2$$.